如何在oracle中根據組中的最小和最大日期獲取值？

Question

我有一個表，其中包含這樣的數據：

車輛	開始	結尾	日期
卡車A	一種	乙	02/01/2021 01:00:00
卡車A	乙	C	02/01/2021 02:00:00
卡車A	C	D	04/01/2021 03:00:00
卡車B	C	一種	05/01/2021 01:00:00
卡車B	一種	乙	06/01/2021 01:00:00
卡車 C	C	乙	07/01/2021 01:00:00
卡車 C	乙	C	08/01/2021 01:00:00
卡車 C	C	乙	09/01/2021 01:00:00
卡車 C	乙	一種	10/01/2021 01:00:00

我需要根據日期獲取每輛車的起點和終點。

例如，卡車 A 在 02/01/2021 01:00:00 從 A 點出發到 B 點。 然后卡車 A 在 02/01/2021 02:00:00 從 B 點移動到 C 點。 卡車 A 已於 04/01/2021 03:00:00 從 C 點移動到 D 點。

它從 A 點開始，在 D 點結束。

我想得到這樣的結果：

車輛	開始	結尾	日期
卡車A	一種	D	04/01/2021 03:00:00

Answer 1

使用GROUP BY並使用KEEP聚合以獲取FIRST或LAST日期值：

SELECT vehicle,
       MIN("START") KEEP (DENSE_RANK FIRST ORDER BY "DATE") AS "START",
       MAX("END")   KEEP (DENSE_RANK LAST  ORDER BY "DATE") AS "END",
       MAX("DATE") AS "DATE"
FROM   table_name
GROUP BY vehicle

其中，對於樣本數據：

CREATE TABLE table_name (Vehicle, "START", "END", "DATE") AS
SELECT 'Truck A', 'A', 'B', DATE '2021-01-02' + INTERVAL '1' HOUR FROM DUAL UNION ALL
SELECT 'Truck A', 'B', 'C', DATE '2021-01-02' + INTERVAL '2' HOUR FROM DUAL UNION ALL
SELECT 'Truck A', 'C', 'D', DATE '2021-01-04' + INTERVAL '3' HOUR FROM DUAL UNION ALL
SELECT 'Truck B', 'C', 'A', DATE '2021-01-05' + INTERVAL '1' HOUR FROM DUAL UNION ALL
SELECT 'Truck B', 'A', 'B', DATE '2021-01-06' + INTERVAL '1' HOUR FROM DUAL UNION ALL
SELECT 'Truck C', 'C', 'B', DATE '2021-01-07' + INTERVAL '1' HOUR FROM DUAL UNION ALL
SELECT 'Truck C', 'B', 'C', DATE '2021-01-08' + INTERVAL '1' HOUR FROM DUAL UNION ALL
SELECT 'Truck C', 'C', 'B', DATE '2021-01-09' + INTERVAL '1' HOUR FROM DUAL UNION ALL
SELECT 'Truck C', 'B', 'A', DATE '2021-01-10' + INTERVAL '1' HOUR FROM DUAL

輸出：

車輛	開始	結尾	日期
卡車A	一種	D	2021-01-04 03:00:00
卡車B	C	乙	2021-01-06 01:00:00
卡車 C	C	一種	2021-01-10 01:00:00

db<> 在這里小提琴

Answer 2

 WITH CTE(Vehicle,  StartT,     EndD,   DateD) AS
 (

     SELECT 'Truck A', 'A' ,'B',    CAST('02/01/2021 01:00:00'AS SMALLDATETIME)  UNION ALL
     SELECT'Truck A' ,  'B',    'C',    CAST('02/01/2021 02:00:00'AS SMALLDATETIME) UNION ALL
     SELECT'Truck A',   'C' ,   'D',    '04/01/2021 03:00:00' UNION ALL
     SELECT'Truck B',   'C' ,   'A',    '05/01/2021 01:00:00' UNION ALL
     SELECT'Truck B',   'A',    'B',    '06/01/2021 01:00:00' UNION ALL
     SELECT'Truck C',   'C' ,   'B',    '07/01/2021 01:00:00' UNION ALL
     SELECT'Truck C' ,  'B',    'C',    '08/01/2021 01:00:00' UNION ALL
     SELECT'Truck C' ,  'C' ,   'B',    '09/01/2021 01:00:00' UNION ALL
     SELECT'Truck C' ,  'B',    'A',    '10/01/2021 01:00:00'  
  ),
 CTE2(Vehicle,  StartT,     EndD,   DateD,XCOL,YCOL) AS
  (
    SELECT C.Vehicle,C.STARTT,C.ENDD,C.DATED,
     ROW_NUMBER()OVER(PARTITION BY C.Vehicle ORDER BY C.DATED ASC)XCOL,
     ROW_NUMBER()OVER(PARTITION BY C.Vehicle ORDER BY C.DATED DESC)YCOL
     FROM CTE AS C
  )
  SELECT C.VEHICLE,C.StartT,X.EndD,X.DateD
   FROM CTE2 AS C
    JOIN CTE2 AS X ON C.Vehicle=X.Vehicle
     WHERE C.XCOL=1 AND X.YCOL=1

抱歉，我沒有 Oracle，因此創建了 MS SQL Server CTE 查詢表示您的示例數據 CTE2 計算“雙向”row_number。 對於起始行（例如對於 'TruckA'）XCOL=1。 對於 end row YCOL=1 然后我們在 Vehicle 上加入這兩個 CTE。 你能試試這個查詢是否適合你

Answer 3

我認為這個解決方案應該適合你，盡管可能有一個更優雅的解決方案。

with x  ( Vehicle   , Start_Letter  ,End_Letter , Date_Letter  )              
as 
(
select 'Truck A' , 'A' , 'B' , to_date('02/01/2021 01:00:00','dd/mm/yyyy hh24:mi:ss') from dual union all 
select 'Truck A' , 'B' , 'C' , to_date('02/01/2021 02:00:00','dd/mm/yyyy hh24:mi:ss') from dual union all 
select 'Truck A' , 'C' , 'D' , to_date('04/01/2021 03:00:00','dd/mm/yyyy hh24:mi:ss') from dual union all 
select 'Truck B' , 'C' , 'A' , to_date('05/01/2021 01:00:00','dd/mm/yyyy hh24:mi:ss') from dual union all 
select 'Truck B' , 'A' , 'B' , to_date('06/01/2021 01:00:00','dd/mm/yyyy hh24:mi:ss') from dual union all 
select 'Truck C' , 'C' , 'B' , to_date('07/01/2021 01:00:00','dd/mm/yyyy hh24:mi:ss') from dual union all 
select 'Truck C' , 'B' , 'C' , to_date('08/01/2021 01:00:00','dd/mm/yyyy hh24:mi:ss') from dual union all 
select 'Truck C' , 'C' , 'B' , to_date('09/01/2021 01:00:00','dd/mm/yyyy hh24:mi:ss') from dual union all 
select 'Truck C' , 'B' , 'A' , to_date('10/01/2021 01:00:00','dd/mm/yyyy hh24:mi:ss') from dual 
) 
select vehicle , start_letter, end_letter, date_letter from 
(
select vehicle , min_start_lett as start_letter, max_start_lett as end_letter, date_letter, 
row_number() over(partition by vehicle order by date_letter desc) as ranking
from 
( 
select x.* , 
rank() over(partition by vehicle order by start_letter desc, end_letter desc, date_letter desc ) as rn , 
min(start_letter) over(partition by vehicle) as min_start_lett , 
max(end_letter) over(partition by vehicle) as max_start_lett
from x 
) d 
where rn = ( select max(rn) from x c where c.vehicle = d.vehicle group by vehicle )             
) where ranking = 1 
;

數據庫<>小提琴

Answer 4

使用窗函數 FIRST_VALUE 查找每輛車的第一個和最后一個

SELECT Vehicle
, FirstStart AS "Start"
, LastEnd AS "End"
, MAX("Date") AS "Date"
FROM
(
  SELECT Vehicle, "Start", "End", "Date"
  , FIRST_VALUE("Start") OVER (PARTITION BY Vehicle ORDER BY "Date") AS FirstStart
  , FIRST_VALUE("End") OVER (PARTITION BY Vehicle ORDER BY "Date" DESC) AS LastEnd
  FROM yourtable
) q
GROUP BY Vehicle, FirstStart, LastEnd
ORDER BY Vehicle;